Solve {c}{frac{a_{2}}{a_{1}}=9}{2a_{1}+3a_{2}=3}

$$\left\{ \begin{array} { c } { \frac { a _ { 2 } } { a _ { 1 } } = 9 } \\ { 2 a _ { 1 } + 3 a _ { 2 } = 3 } \end{array} \right.$$

Solve for a_2, a_1

$$a_{2}=\frac{27}{29}\approx 0.931034483$$

$$a_{1}=\frac{3}{29}\approx 0.103448276$$

Steps Using Substitution

Consider the first equation. Variable a_{1} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a_{1}.

$$a_{2}=9a_{1}$$

Substitute 9a_{1} for a_{2} in the other equation, 3a_{2}+2a_{1}=3.

$$3\times 9a_{1}+2a_{1}=3$$

Multiply 3 times 9a_{1}.

$$27a_{1}+2a_{1}=3$$

Add 27a_{1} to 2a_{1}.

$$29a_{1}=3$$

Divide both sides by 29.

$$a_{1}=\frac{3}{29}$$

Substitute \frac{3}{29} for a_{1} in a_{2}=9a_{1}. Because the resulting equation contains only one variable, you can solve for a_{2} directly.

$$a_{2}=9\times \frac{3}{29}$$

Multiply 9 times \frac{3}{29}.

$$a_{2}=\frac{27}{29}$$

The system is now solved.

$$a_{2}=\frac{27}{29},a_{1}=\frac{3}{29}$$

Steps Using Matrices

Consider the first equation. Variable a_{1} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a_{1}.

$$a_{2}=9a_{1}$$

Subtract 9a_{1} from both sides.

$$a_{2}-9a_{1}=0$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$a_{2}-9a_{1}=0,3a_{2}+2a_{1}=3$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-9\\3&2\end{matrix}\right)\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=\left(\begin{matrix}0\\3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-9\\3&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-9\\3&2\end{matrix}\right))\left(\begin{matrix}1&-9\\3&2\end{matrix}\right)\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-9\\3&2\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-9\\3&2\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-9\\3&2\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2-\left(-9\times 3\right)}&-\frac{-9}{2-\left(-9\times 3\right)}\\-\frac{3}{2-\left(-9\times 3\right)}&\frac{1}{2-\left(-9\times 3\right)}\end{matrix}\right)\left(\begin{matrix}0\\3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{2}{29}&\frac{9}{29}\\-\frac{3}{29}&\frac{1}{29}\end{matrix}\right)\left(\begin{matrix}0\\3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{9}{29}\times 3\\\frac{1}{29}\times 3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{2}\\a_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{27}{29}\\\frac{3}{29}\end{matrix}\right)$$

Extract the matrix elements a_{2} and a_{1}.

$$a_{2}=\frac{27}{29},a_{1}=\frac{3}{29}$$

Steps Using Elimination

Consider the first equation. Variable a_{1} cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a_{1}.

$$a_{2}=9a_{1}$$

Subtract 9a_{1} from both sides.

$$a_{2}-9a_{1}=0$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$a_{2}-9a_{1}=0,3a_{2}+2a_{1}=3$$

To make a_{2} and 3a_{2} equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.

$$3a_{2}+3\left(-9\right)a_{1}=0,3a_{2}+2a_{1}=3$$

Simplify.

$$3a_{2}-27a_{1}=0,3a_{2}+2a_{1}=3$$

Subtract 3a_{2}+2a_{1}=3 from 3a_{2}-27a_{1}=0 by subtracting like terms on each side of the equal sign.

$$3a_{2}-3a_{2}-27a_{1}-2a_{1}=-3$$

Add 3a_{2} to -3a_{2}. Terms 3a_{2} and -3a_{2} cancel out, leaving an equation with only one variable that can be solved.

$$-27a_{1}-2a_{1}=-3$$

Add -27a_{1} to -2a_{1}.

$$-29a_{1}=-3$$

Divide both sides by -29.

$$a_{1}=\frac{3}{29}$$

Substitute \frac{3}{29} for a_{1} in 3a_{2}+2a_{1}=3. Because the resulting equation contains only one variable, you can solve for a_{2} directly.

$$3a_{2}+2\times \frac{3}{29}=3$$

Multiply 2 times \frac{3}{29}.

$$3a_{2}+\frac{6}{29}=3$$

Subtract \frac{6}{29} from both sides of the equation.

$$3a_{2}=\frac{81}{29}$$

Divide both sides by 3.

$$a_{2}=\frac{27}{29}$$

The system is now solved.

$$a_{2}=\frac{27}{29},a_{1}=\frac{3}{29}$$